Question

5. If x2 + k(4x + k-1) + 2 = 0 has equal roots, then k=​

Answer 1

Answer:

1+4272+76529+8182+2(71992+5272902?62299202+272901911)2792////272029!!!!

Answer 2

hey buddy here is your answer

Given : x² + k(4x + k - 1)  + 2 = 0

x² + 4kx + k² - k + 2 = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 1 , b =  4k  , c = k² - k + 2

D(discriminant) = b² – 4ac  

D = (4k)² - 4 × 1 × (k² - k + 2)  

D = 16k² - 4k² + 4k - 8

D = 12k² + 4k - 8

D = 4(3k² + k - 2)

D = 0 (equal roots given)

4(3k² + k - 2) = 0

(3k² + k - 2) = 0

3k² + 3k - 2k - 2 = 0

[By middle term splitting]

3k (k + 1) - 2(k + 1) = 0  

(3k - 2)(k + 1) = 0

(3k - 2) = 0 (k + 1) = 0

3k = 2 or k = - 1

k = ⅔ or k = - 1

Hence, the value of k is ⅔ & - 1.

Hope it helps buddy

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